Question

If \[ f(x) = 2 \left( \cos x + i \sin x \right) \left( \cos 3x + i \sin 3x \right) \cdots \left( \cos (2n-1)x + i \sin (2n-1)x \right) \] where \( n \in \mathbb{N} \), then what is the value of \( f''(x) \) ?

• A. \( -n^2 f(x) \)
• B. \( n^2 f(x) \)
• C. \( -n^4 f(x) \)
• D. \( n^4 f(x) \)

Hint:

Products of exponential functions like \( e^{ikx} \) can be simplified using summation properties. Also, recall the sum of first \( n \) odd numbers is \( n^2 \). Differentiation of \( e^{ax} \) follows the chain rule, bringing down powers accordingly.

Answer 1

The Correct Option is C

We are given a function written as a product of complex exponential forms. Using Euler's formula: \[ \cos kx + i \sin kx = e^{ikx} \] So, \[ f(x) = 2 \cdot e^{ix} \cdot e^{i3x} \cdot e^{i5x} \cdots e^{i(2n-1)x} \] There are \( n \) such terms: \( 1, 3, 5, ..., (2n-1) \), which are the first \( n \) odd numbers. So, \[ f(x) = 2 \cdot \exp\left[i(1 + 3 + 5 + \dots + (2n-1))x\right] \] Now, we know that: \[ 1 + 3 + 5 + \dots + (2n-1) = n^2 \] Therefore: \[ f(x) = 2 \cdot e^{i n^2 x} \] Now compute \( f''(x) \): First derivative: \[ f'(x) = 2 \cdot i n^2 e^{i n^2 x} \] Second derivative: \[ f''(x) = 2 \cdot (i n^2)^2 e^{i n^2 x} = 2 \cdot (-n^4) e^{i n^2 x} = -n^4 f(x) \] So the final answer is: \[ f''(x) = -n^4 f(x) \]